Giúp mị với mọi ngừi ơi:(((((((

Giải thích các bước giải:

 ĐK: $x,y,z>0$

Ta có:

$\begin{array}{l} x + y + z + xy + yz + xz = 6xyz\\  \Leftrightarrow \dfrac{{x + y + z + xy + yz + xz}}{{xyz}} = 6\\  \Leftrightarrow \dfrac{1}{{yz}} + \dfrac{1}{{xz}} + \dfrac{1}{{xy}} + \dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y} = 6 \end{array}$

Mà lại có:

$\begin{array}{l} A = \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}\\  \Rightarrow 3A + 3 = \left( {\dfrac{1}{{{x^2}}} + 1} \right) + \left( {\dfrac{1}{{{y^2}}} + 1} \right) + \left( {\dfrac{1}{{{z^2}}} + 1} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right) + \left( {\dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{z^2}}}} \right)\\  \Rightarrow 3A + 3 \ge 2\sqrt {\dfrac{1}{{{x^2}}}}  + 2\sqrt {\dfrac{1}{{{y^2}}}}  + 2\sqrt {\dfrac{1}{{{z^2}}}}  + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{y^2}}}}  + 2\sqrt {\dfrac{1}{{{y^2}}}.\dfrac{1}{{{z^2}}}}  + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{z^2}}}} \\  \Rightarrow 3A + 3 \ge 2.\dfrac{1}{x} + 2.\dfrac{1}{y} + 2.\dfrac{1}{z} + 2.\dfrac{1}{{xy}} + 2.\dfrac{1}{{yz}} + 2.\dfrac{1}{{xz}}\\  \Rightarrow 3A + 3 \ge 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} + \dfrac{1}{{xy}} + \dfrac{1}{{yz}} + \dfrac{1}{{xz}}} \right)\\  \Rightarrow 3A + 3 \ge 2.6\\  \Rightarrow A \ge 3 \end{array}$

Dấu bằng xảy ra

$\Leftrightarrow \left\{ \begin{array}{l} \dfrac{1}{x} = 1\\ \dfrac{1}{y} = 1\\ \dfrac{1}{z} = 1\\ \dfrac{1}{x} = \dfrac{1}{y}\\ \dfrac{1}{y} = \dfrac{1}{z}\\ \dfrac{1}{x} = \dfrac{1}{z} \end{array} \right. \Leftrightarrow x = y = z = 1$

Vậy ta có điều phải chứng minh.